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 January 15th, 2013, 07:14 AM #1 Newbie   Joined: Jan 2013 Posts: 3 Thanks: 0 Lagrange multiplier method So I solved a problem, but don't really know the purpose... Could it be that these to points are the maximum points or am I missing something?
 January 15th, 2013, 10:29 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Lagrange multiplier method We are given the objective function: $U(x,y)=100-e^{-x}-e^{-y}$ subject to the constraint: $g(x,y)=px+qy-m=0$ The method of Lagrange multipliers then gives us: $e^{-x}=\lambda p$ $e^{-y}=\lambda q$ Now this implies: $\lambda=\frac{e^{-x}}{p}=\frac{e^{-y}}{q}$ $pe^x=qe^y$ $y=x+\ln$$\frac{p}{q}$$$ Substituting this into the constraint, we find: $px+q$$x+\ln\(\frac{p}{q}$$\)-m=0$ $(p+q)x=m-q\ln$$\frac{p}{q}$$$ $x=\frac{m-q\ln$$\frac{p}{q}$$}{p+q}$ and so: $y=\frac{m-q\ln$$\frac{p}{q}$$}{p+q}+\ln$$\frac{p}{q}$$=\frac {m+p\ln$$\frac{p}{q}$$}{p+q}$ Although in slightly different form, my result for the critical point agrees with yours. This is the point which maximizes the objective function, or alternately, which minimizes $e^{-x}+e^{-y}$ subject to the given constraint.

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### lagrange multiplier method

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